Question

A balloon which always remains spherical is being inflated by pumping in $10$ cubic centimeters of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is $15cms$

• A. $\frac{1}{90\pi }cm/sec$
• B. $\frac{1}{9\pi }cm/sec$
• C. $\frac{1}{30\pi }cm/sec$
• D. $\frac{1}{\pi }cm/sec$

Answer 1

The correct option is A $\frac{1}{90\pi }cm/sec$

The volume of a sphere $V$ with radius $r$ is given by,

$$\begin{array}{c}V=\frac{4}{3}\pi r^3\end{array}$$

∴Rate of change of volume $V$ with respect to time $t$ is given by,

$$\begin{array}{c}\frac{dV}{dt}=\frac{dV}{dr}.\frac{dr}{dt}\left[ByChainRule\right]\end{array}$$

$$\begin{array}{c}=\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right).\frac{dr}{dt}\end{array}$$

$$\begin{array}{c}=4\pi r^2.\frac{dr}{dt}\end{array}$$

It is given that

$$\begin{array}{c}\frac{dV}{dt}=10c{m}^3/s\end{array}$$

$$\begin{array}{c}\therefore 10=4\pi r^2.\frac{dr}{dt}\end{array}$$

$$\begin{array}{c}\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}=\frac{5}{2\pi r^2}\end{array}$$

Therefore, when radius $=15$ cm,

$$\begin{array}{c}\frac{dr}{dt}=\frac{5}{2\pi (15{)}^2}=\frac{1}{90\pi }\end{array}$$

Hence, the rate at which the radius of the balloon increases when the radius is $15$ cm is $\frac{1}{90\pi }cm/sec$